Question 1182506
<pre>
when the degree is 1, P(x) = c<sub>1</sub>x+c<sub>2</sub>, certainly you can
put in infinitely many positive integers for x and get a composite number.

For P(x) with degrees m > 1:

{{{P(x)}}}{{{""=""}}}{{{sum(c[k](x^k),k=0,m)}}}, c<sub>i</sub> ≥ 0.

{{{P(a)-P(b)}}}{{{""=""}}}{{{sum(c[k](a^k),k-0,n)-sum(c[k](b^k),k=0,m)}}}{{{""=""}}}{{{sum(c[k](a^k-b^k),k=0,m)}}}{{{""=""}}}

{{{sum(c[k](a-b)sum(a^(k-1-j)*b^j,j=0,k-1) ,k=0,m)}}}{{{""=""}}}{{{(a-b)sum(c[k]sum(a^(k-1-j)*b^j,j=0,k-1) ,k=0,m)}}}{{{""=""}}}

{{{(a-b)(matrix(1,8,the,sum,of,two, or, more,positive,integers))}}}

There are an infinite number of ways we can choose a and b a > b, such
that a-b is not 1.  And the summation is of two or more positive integers,
which are integers > 1. 

So there exist infinitely many positive integers n such that P(n) is
composite.

Edwin</pre>